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waltersBest ResponseYou've already chosen the best response.0
jjj let the mapping \[\tau _{ab}\] for a,b element R ,maps the reals by the ruel \[\tau _{ab}:x \rightarrow ax+b.Let G={\tau _{ab}:a \neq 0}\] Determine whether or not G is an abelian group under the composition of mappings
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.4
Well, let \(\tau_{ab},\tau_{cd}\in G\). Then \[\tau_{ab}\circ\tau_{cd}(x)=\tau_{ab}(cx+d)=acx+ad+b.\]Also, \[\tau_{cd}\circ\tau_{ab}(x)=\tau_{cd}(ax+b)=cax+bc+d.\]Since \(acx+ad+b\neq acx+bc+d\), it must be that \(G\) is not abelian.
 one year ago

waltersBest ResponseYou've already chosen the best response.0
am i not suppose to check whether is a group using the condition before i can conclude that is an abelian group
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.4
It just asked if it was an abelian group. Since the elements don't commute, it can't be an abelian group even if it is a group. However, if we want to check it's a group, we can do that as well. Have you shown any of the axioms for proving it's a group yet?
 one year ago

waltersBest ResponseYou've already chosen the best response.0
i am failing to show the axioms
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.4
First, we can see that \(G\) is closed under composition. \[\tau_{ab}\circ\tau_{cd}(x)=\tau_{ab}(cx+d)=acx+ad+b=\tau_{(ac)(ad+b)}.\]We also have the identity \(\tau_{10}\). Now we just have to check associativity and inverses.
 one year ago

waltersBest ResponseYou've already chosen the best response.0
why do u chose to use dw:1360875075213:dw
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.4
"\(\circ\)" is a relatively common notation to denote the composition of two functions. You can also use \(\tau_{ab}(\tau_{cd}(x))\) is you prefer.
 one year ago

waltersBest ResponseYou've already chosen the best response.0
can u also use * if u wnt
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.4
For inverses, we want \[\tau_{ab}\circ\tau_{cd}(x)=acx+ad+b=x\]for some \(c,d\in\mathbb{R}\). We can see that \(c=a^{1}\) and \(d=a^{1}(b)\) work for this direction. For the other, \[\tau_{cd}\circ\tau_{ab}(x)=cax+bc+d=a^{1}ax+ba^{1}a^{1}b=x,\]so we do indeed have inverses.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.4
If you make it clear your group function is composition of functions, you can use * as well.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.4
As for associativity, I'm just going to use the fact that composition of functions is always associative. If you have not heard of that yet, we can still go over how to prove this particular function is associative.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.4
let \(\tau_{ab},\tau_{cd},\tau_{ef}\in G\). Then\[\tau_{ab}\circ(\tau_{cd}\circ\tau_{ef})(x)=\tau_{ab}=\tau_{ab}(cex+cf+d)=acex+acf+ad+b\]and\[(\tau_{ab}\circ\tau_{cd})\circ\tau_{ef}(x)=(\tau_{(ac)(ad+b)})(ex+f)=acex+acf+ad+b\]Since these are equal, it's associative, and we have a group.
 one year ago

waltersBest ResponseYou've already chosen the best response.0
k it is a group but not an abelian group
 one year ago

waltersBest ResponseYou've already chosen the best response.0
can we verify identity
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.4
\[\tau_{10}(x)=1x+0=x\]
 one year ago
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